AtCoder Beginner Contest 374

A - Takahashi san 2

思路

取末三位的子串判断即可；

AC代码

    #include<bits/stdc++.h>
    #define endl '\n'
    #define int int long long
    #define pb push_back
    #define bs bitset
    using namespace std;
    typedef pair<char,int> PCI;
    typedef pair<int,int> PII;
    typedef priority_queue<int> PQ;

    const int N = 2e5+10, MAX = 1e9, INF = -1e9;

    string s;

    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        cin>>s;
        int l=s.size();
        if(s.substr(l-3)=="san")cout<<"Yes"<<endl;
        else cout<<"No"<<endl;  


        return 0;
    }

B - Unvarnished Report

思路

判断 \(S\) 和 \(T\) 相同输出“Yes”，否则找出不同位置即可（注意别越界）；

AC代码

    #include<bits/stdc++.h>
    #define endl '\n'
    #define int int long long
    #define pb push_back
    #define bs bitset
    using namespace std;
    typedef pair<char,int> PCI;
    typedef pair<int,int> PII;
    typedef priority_queue<int> PQ;

    const int N = 2e5+10, MAX = 1e9, INF = -1e9;

    string s,t;

    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        cin>>s>>t;
        if(s==t){
            cout<<"0"<<endl;
            return 0;
        }
        int ls=s.size();
        int lt=t.size();
        if(ls>=lt){
            s+="X";
            for(int i=0;i<ls;i++){
                if(s[i]!=t[i]){
                    cout<<i+1<<endl;
                    return 0;
                }
            }
        }
        else{
            t+="X";
            for(int i=0;i<lt;i++){
                if(s[i]!=t[i]){
                    cout<<i+1<<endl;
                    return 0;
                }
            }
        }

        return 0;
    }

C - Separated Lunch

思路

\(N\) 的范围较小，直接DFS即可；

AC代码

    #include<bits/stdc++.h>
    #define endl '\n'
    #define int int long long
    #define pb push_back
    #define bs bitset
    using namespace std;
    typedef pair<char,int> PCI;
    typedef pair<int,int> PII;
    typedef priority_queue<int> PQ;

    const int N = 25, MAX = 1e9, INF = -1e9;

    int n;
    int a[N];
    int ans=MAX;

    void dfs(int x,int l,int r){
        if(x>n){
            ans=min(ans,max(l,r));
            return ;
        }
        dfs(x+1,l+a[x],r);
        dfs(x+1,l,r+a[x]);
    }

    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        cin>>n;
        for(int i=1;i<=n;i++){
            cin>>a[i];
        }
        dfs(1,0,0);
        cout<<ans<<endl;
        return 0;
    }

D - Laser Marking

思路

1.把{0,0}点看作第0个线段，长度为0;
2.生成1~n的全排列，即线段的打印顺序;
3.对于每个排列dfs时间，key表示顺序打印和逆序打印;
4.对于每个线段，加上打印时间，从线段末端点转移到下一个端点（dfs）;
时间复杂度大致为：\(O(n! 2^n)\)

AC代码

    #include <bits/stdc++.h>
    #define endl '\n'
    #define int long long
    #define pb push_back
    #define bs bitset
    using namespace std;

    typedef pair<double, double> PDD;

    const int N = 8;
    const double INF = 1e9; 

    int n, t, ss;
    vector<PDD> s(N);
    vector<PDD> e(N);
    double len[N];
    double ans = INF;
    vector<int> ord(N);

    void dfs(int x, bool key, double l) {
        l += len[ord[x]] / t;
        if (x == n) { 
            ans = min(ans, l);
            //cout<<l<<endl;
            return;
        }

        double x0, y0;
        if (key) {
            x0 = e[ord[x]].first; 
            y0 = e[ord[x]].second;
        } else {
            x0 = s[ord[x]].first; 
            y0 = s[ord[x]].second;
        }

        double l1 = sqrtl((x0 - s[ord[x + 1]].first) * (x0 - s[ord[x + 1]].first) +
                        (y0 - s[ord[x + 1]].second) * (y0 - s[ord[x + 1]].second))/ss ;

        double l0 = sqrtl((x0 - e[ord[x + 1]].first) * (x0 - e[ord[x + 1]].first) +
                        (y0 - e[ord[x + 1]].second) * (y0 - e[ord[x + 1]].second))/ss;

        dfs(x + 1, 1, l + l1); 
        dfs(x + 1, 0, l + l0); 
    }

    signed main() {
        ios::sync_with_stdio(false);
        cin.tie(0);
        
        cin >> n;
        cin >> ss >> t;
        s[0]={0,0};e[0]={0,0};len[0]=0.0;

        for (int i = 1; i <= n; i++) { 
            cin >> s[i].first >> s[i].second;
            cin >> e[i].first >> e[i].second;
            len[i] = sqrtl((s[i].first - e[i].first) * (s[i].first - e[i].first) + 
                        (s[i].second - e[i].second) * (s[i].second - e[i].second));
        }
        
        for (int i = 1; i <= n; i++) {
            ord[i] = i; 
        }
        
        do {
            dfs(0, true, 0); 
        } while (next_permutation(ord.begin()+1, ord.begin() + 1 + n)); 

        cout << fixed << setprecision(12) << ans << endl; 

        return 0;
    }

E - Sensor Optimization Dilemma 2

思路

目标是寻找 \(W\) 最小值中的最大值，考虑二分；二分模板中我们只需要编写 \(check()\) 函数即可，我们需要检查当前花费条件下是否可以达到 \(W\) 的生产效率，由于 \(N\) 较小，考虑枚举，每次优先选择性价比较高的机器，再逐步替换，找到最小花费，与 \(x\) 比较返回,时间复杂度大致为\(O(nlogx)\)；

AC代码

#include<bits/stdc++.h>
#define endl '\n'
#define int int long long
#define pb push_back
#define bs bitset
using namespace std;
typedef pair<char,int> PCI;
typedef pair<int,int> PII;
typedef priority_queue<int> PQ;
const int N = 2e5+10, MAX = 1e9, INF = -1e9;

int n,x;
int A[N], B[N], P[N], Q[N];

bool check(int w){
    int sum=0;
    for(int i=1;i<=n;i++){
        int a = A[i], b = B[i], p = P[i], q = Q[i];
        if(a * q < b * p) {
            swap(a,b);swap(p,q);
        }
        int res = 1e18;
            for(int j = 0;j < 100;j ++) {
                int v = j * q;
                int need = w - j * b;
                v += (need + a - 1) / a * p;
                res = min(res, v);
            }
        sum += res;
    }
    return sum<=x;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> x;
	for (int i = 1; i <= n; i++)
	{
		cin >> A[i] >> P[i] >> B[i] >> Q[i];
	}
	int l = 0, r = 1e9;
	while (l < r)
	{
		int mid = l + r + 1 >> 1;
		if (check(mid))
		{
			l = mid;
		}
		else
			r = mid - 1;
	}
	cout << l << endl;

    return 0;
}

t.b.c.